How to know if a list has an even or odd number of elements

Question

How can I find out if there is even, or odd, number of elements in an arbitrary list.

I tried list.index() to get all of the indices... but I still don\'t k

Answer 1

You can use the built in function len() for this.

Python Doc -- len()

Gets the length (# of elements) of any arbitrary list.

myList = [0,1,2,3,4,5]

if len(myList) % 2 == 0:
    print ("even")
else
    print ("odd")

Define function that returns a bool (true or false).

def is_even(myList):

    if len(myList) % 2 == 0:
        return true
    else:
        return false

main():

    myList = [0,1,2,3]
    theListIsEven = is_even(myList)  # will be true in this example
                                     # because 4 items in myList

    if theListIsEven(myList) == True:
        # do something
    else:
        # do something else

    return 0

The modulus operator % gives the remainder.

EX: 7 % 2 = 1

• Closest number to 7 that 2 will divide evenly is 6
• Which is 1 away from 7.
• Thus, remainder of 1 for 7 % 2.

EX: 4 % 2 = 0

• Any even number n will give 0 as the remainder when n % 2
• Because n has divided evenly by 2

Answer 2

def has_even_length(some_sequence):
    return not len(some_sequence)%2

def has_odd_length(some_sequence):
    return bool(len(some_sequence)%2)

Answer 3

your_list = [1,2,3,(4,5)]

# modulo operation finds the remainder of division of one number by another.
if len(your_list) % 2 == 0:
    print "Even Number"
else:
    print"number is odd"

Answer 4

Even numbers are divisible by 2. Odd numbers are not.

len(X) will get the length of X

If the length of X is divisible by 2, then it is an Even number If the length of X is not divisible by 2 then it is an Odd Number

len(X)%2 returns the "remainder" of a division problem

for example 5%2 will return 1 which is NOT zero, (because 5 divided by 2 is 2 with a remainder of 1), therefore it is not even. Same thing as 6%4 which would return a 2, because 6 divided by 4 is 1 with a remainder of 2.

so len(X)%2 where X is your list, will return either a 1, indicating it is Odd, or a 0 indicating it is Even.

Answer 5

if you want to be bitwise about it you could also use.

if len(mylist)&1:
    print("odd")
else:
    print("even")

this is really Nice looking if you know your list is more likely to be odd.

if len(mylist)&1:
    print("list was odd")
elif CheckSomthingElse():
    print("list was even AND CheckSomthingElse was True")
else:
    print("list was even AND CheckSomthingElse was False")

Wow in python by the test below % 2 is faster. python must be optimized for it. in some other languages & 1 is faster. this is why it's important to always test.

import timeit

MyListOdd=[1,2,3,4,5,6,7]
MyListEven=[1,2,3,4,5,6]
print("MyListOdd  == ",MyListOdd)
print("MyListEven == ",MyListEven)
print("len(MyListOdd)  == ",len(MyListOdd))
print("len(MyListEven) == ",len(MyListEven))

def TestIfEvenBitwise(MyList):
    if len(MyList)&1:
        return False #Odd
    else:
        return True #Even

def TestIfEvenModulus(MyList):
    if len(MyList)%2:
        return False #Odd
    else:
        return True #Even

print("TestIfEvenBitwise(MyListOdd)  == ",TestIfEvenBitwise(MyListOdd))
print("TestIfEvenModulus(MyListOdd)  == ",TestIfEvenModulus(MyListOdd))
print("TestIfEvenBitwise(MyListEven) == ",TestIfEvenBitwise(MyListEven))
print("TestIfEvenModulus(MyListEven) == ",TestIfEvenModulus(MyListEven))

mysetup = """
MyListOdd=[1,2,3,4,5,6,7]
MyListEven=[1,2,3,4,5,6]

def TestIfEvenBitwise(MyList):
    if len(MyList)&1:
        return False #Odd
    else:
        return True #Even

def TestIfEvenModulus(MyList):
    if len(MyList)%2:
        return False #Odd
    else:
        return True #Even
"""

print("timeit.timeit(setup = mysetup,stmt ='TestIfEvenBitwise(MyListOdd)', number=100000) == ",
      timeit.timeit(setup = mysetup,stmt ='TestIfEvenBitwise(MyListOdd)', number=100000))
print("timeit.timeit(setup = mysetup,stmt ='TestIfEvenModulus(MyListOdd)', number=100000) == ",
      timeit.timeit(setup = mysetup,stmt ='TestIfEvenModulus(MyListOdd)', number=100000))
print("timeit.timeit(setup = mysetup,stmt ='TestIfEvenBitwise(MyListEven)', number=100000) == ",
      timeit.timeit(setup = mysetup,stmt ='TestIfEvenBitwise(MyListEven)', number=100000))
print("timeit.timeit(setup = mysetup,stmt ='TestIfEvenModulus(MyListEven)', number=100000) == ",
      timeit.timeit(setup = mysetup,stmt ='TestIfEvenModulus(MyListEven)', number=100000))

results of test.

MyListOdd  ==  [1, 2, 3, 4, 5, 6, 7]
MyListEven ==  [1, 2, 3, 4, 5, 6]
len(MyListOdd)  ==  7
len(MyListEven) ==  6
TestIfEvenBitwise(MyListOdd)  ==  False
TestIfEvenModulus(MyListOdd)  ==  False
TestIfEvenBitwise(MyListEven) ==  True
TestIfEvenModulus(MyListEven) ==  True
timeit.timeit(setup = mysetup,stmt ='TestIfEvenBitwise(MyListOdd)', number=100000) ==  0.02574796500000004
timeit.timeit(setup = mysetup,stmt ='TestIfEvenModulus(MyListOdd)', number=100000) ==  0.022446242000000005
timeit.timeit(setup = mysetup,stmt ='TestIfEvenBitwise(MyListEven)', number=100000) ==  0.026081517000000054
timeit.timeit(setup = mysetup,stmt ='TestIfEvenModulus(MyListEven)', number=100000) ==  0.025758655999999935

Answer 6

All you need is

len(listName)

Which will give you the length.

I guess you could also do this then

if len(listName) % 2 == 0:
    return True  # the number is even!
else:
    return False # the number is odd!